2 Answers
2

Markus Brede proves the following formula in the paper "On the convergence of the sequence defining Euler’s number". Let $$\left(1+\frac{1}{z}\right)^z=\sum_{n\geq 0} \frac{a_n}{z^n}$$
then we have
$$a_n=e\sum_{v=0}^n \frac{S(n+v,v)}{(n+v)!}\sum_{m=0}^{n-v}\frac{(-1)^m}{m!}$$
where $S(a,b)$ are Stirling numbers of the first kind. This shows that all coefficients are rational multiples of $e$. I found the article through OEIS.

Thank you, that is excellent. I should have gone straight to OEIS. I can't believe that I didn't think to do so. It turns out that I also needed another related series which was easily found there.
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ChrisOct 7 '11 at 9:52

$\log f(z) = z \log(1+1/z) = \sum_{k=0}^\infty \frac{(-1)^k}{k+1} z^{-k}$ as $z \to +\infty$, so $f(z)$ is the exponential of this sum. See http://oeis.org/A055505 for the numerators and http://oeis.org/A055535 for the denominators of the coefficients.